## Calculus (3rd Edition)

$$\frac{1}{3} \leq \int_{0}^{1} \frac{d x}{\sqrt{5 x^{3}+4}} \leq \frac{1}{2}$$
Since$$f^{\prime}(x)=\frac{15 * x^{2}}{\sqrt{5 x^{3}+4}} \geq 0$$ then $f(x)$ is increasing on $[ 0,1]$ and $\int_{0}^{1} \frac{d x}{\sqrt{5 x^{3}+4}}$ has an upper bound at $x=0$ $$\int_{0}^{1} \frac{1}{\sqrt{5(0)+4}}=\int_{0}^{1} \frac{1}{2} d x=1 / 2$$ and a lower bound $$\int_{0}^{1} \frac{1}{\sqrt{5(1)+4}}=\int_{0}^{1} \frac{1}{3} d x=1 / 3$$ Then, by using the comparison theorem \begin{aligned} \int_{0}^{1} \frac{1}{3} d x & \leq \int_{0}^{1} \frac{d x}{\sqrt{5 x^{3}+4}} \leq \int_{0}^{1} \frac{1}{2} d x \\ \frac{1}{3} *(1-0) & \leq \int_{0}^{1} \frac{d x}{\sqrt{5 x^{3}+4}} \leq \frac{1}{2} *(1-0) \\ \frac{1}{3} & \leq \int_{0}^{1} \frac{d x}{\sqrt{5 x^{3}+4}} \leq \frac{1}{2} \end{aligned}